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### Stability and determinacy A structure is externally stable only if its support reactions prevent translation in two independent directions and rotation. It is statically determinate when all reaction and internal forces follow from equilibrium alone; otherwise, the number of extra force unknowns is its degree of indeterminacy. 1. In FE-TOS-002, the support reaction directions do not provide three independent restraints for the assembly. A rigid-body motion is possible, so it is $\boxed{\text{externally unstable}}$. 2. In FE-TOS-003, separating the structure at the internal hinge and counting the independent equilibrium equations leaves one redundant reaction. It is $\boxed{\text{indeterminate to }1^\circ}$. 3. In FE-TOS-004, the support reaction arrangement permits a rigid-body motion of the frame. Thus it is $\boxed{\text{externally unstable}}$.

Counting rules Use: $$ I=3b+r-3j-c, $$ where $b$ is the number of bars, $r$ is the number of support reactions, $j$ is the number of joints, and $c$ is the number of equations of condition. Bar ends, connections, and changes in direction count as joints. A change in direction creates a new bar for counting, even if the corner is rigid. For an internal connection between two members: An internal hinge transmits force but no bending moment. Its zero-moment condition gives one additional equation: $$ M=0 \qquad\Rightarrow\qquad c=1. $$ An internal roller transmits only force normal to its rolling surface. Both the bending moment and the force parallel to that surface are zero, giving two additional equations: $$ M=0,\qquad F_{\parallel}=0 \qquad\Rightarrow\qquad c=2. $$ These values apply to internal connections. External supports are counted through $r$. 1. Frame 1 There are four bars, five joints, three support reactions, and one internal hinge. $$ b=4,\qquad r=3,\qquad j=5,\qquad c=1. $$ $$ I=3(4)+3-3(5)-1=-1. $$ The negative result indicates insufficient constraints. $$ \boxed{\text{Frame 1 is unstable.}} $$ 2. Frame 2 Taking the upright and inclined members as continuous, with the two horizontal bars pin-connected at both ends, there are six bars, six joints, three support reactions, and four equations of condition. $$ b=6,\qquad r=3,\qquad j=6,\qquad c=4. $$ $$ I=3(6)+3-3(6)-4=-1. $$ $$ \boxed{\text{Frame 2 is unstable under this interpretation.}} $$ The highlighted “determinate” answer conflicts with this reading of the drawing. If the right support were a pin instead of a roller, then $r=4$: $$ I=3(6)+4-3(6)-4=0. $$ With that change, the frame would be stable and determinate. 3. Frame 3 The rigid elbow creates a joint and divides the frame into two bars. The fixed support provides three reactions. $$ b=2,\qquad r=3,\qquad j=3,\qquad c=0. $$ $$ I=3(2)+3-3(3)-0=0. $$ The frame is stable. $$ \boxed{\text{Frame 3 is statically determinate.}} $$ 4. Frame 4 The middle connections divide the columns into four bars. Including the two beams gives six bars. The two fixed supports provide six reactions. $$ b=6,\qquad r=6,\qquad j=6,\qquad c=0. $$ $$ I=3(6)+6-3(6)-0=6. $$ The frame is stable. $$ \boxed{\text{Frame 4 is statically indeterminate to the sixth degree.}} $$

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Figure 1

Each fixed support supplies three reaction components. The two frictionless rollers between the beams each transmit one vertical contact force.

Separate the assembly into two beam bodies. Each roller force appears as an equal-and-opposite pair but represents only one independent unknown. There are eight unknowns and six independent equilibrium equations.

$$I = [2(3) + 2] - 3(2)$$ $$I = 8 - 6 = 2$$

Stability: Each beam is independently stable as a fixed-ended cantilever. The roller contacts introduce two redundant constraints without creating a mechanism, assuming both contacts remain engaged.

Answer: $I = 2$. The structure is stable and statically indeterminate to the second degree.

Figure 2

The bent member is one continuous rigid body without an internal hinge at its elbow. Each frictionless roller provides one reaction normal to its supporting surface.

$$r = 1 + 1 + 1 = 3$$

A planar rigid body has three equilibrium equations.

$$I = r - 3$$ $$I = 3 - 3 = 0$$

Stability: The reactions at A and B are parallel but act along distinct lines. The reaction at C is not parallel to them. Therefore, the three reactions are neither all parallel nor all concurrent.

The reaction lines at A and C intersect at O. The reaction at B does not pass through O, so it has a nonzero moment arm about O and restrains rotation. Together, the supports also restrain translation.

The continuous rigid member has no internal mechanism.

Answer: $I = 0$. The structure is stable and statically determinate.

Figure 3

The crossing members are not connected at their intersections. Therefore, the crossings add neither joints nor member subdivisions.

There are six actual joints: A, B, C, D, E, and F.

$$j = 6$$

Count four chord members (AB, BC, DE, EF), two end vertical members (AD, CF), and four diagonal members (DB, EC, AE, BF).

$$m = 4 + 2 + 4 = 10$$

The pin at A supplies two reaction components, and the roller at C supplies one.

$$r = 2 + 1 = 3$$

Apply the plane-truss equation:

$$I = m + r - 2j$$ $$I = 10 + 3 - 2(6)$$ $$I = 1$$

Stability: The pin-and-roller supports restrain rigid-body translation and rotation. The diagonal arrangement prevents internal distortion even though the crossings are not joints.

The external reactions can be determined from the three overall equilibrium equations. Therefore, the single redundancy is internal.

Answer: $I = 1$. The truss is stable and statically indeterminate to the first degree, internally.

Instability Classification

None of the three structures is unstable under the stated idealizations.

External instability: The entire structure can translate or rotate because its supports provide inadequate restraint, such as when all reaction lines are parallel or concurrent.

Internal instability: The structure can change shape through an internal mechanism even when its supports restrain rigid-body motion.

For plane trusses, $I = m + r - 2j$ is a counting check. A negative value indicates insufficient constraints. A zero or positive value still requires checking the support directions and member arrangement for mechanisms.